What happens to the capacitance when a dielectric fully fills the space between the plates?

What happens to the capacitance when a dielectric fully fills the space between the plates?

• A. C' = κ C0
• B. C' = C0 / κ
• C. C' = κ^2 C0
• D. C' = C0

Answer: (A)

The key idea is that inserting a dielectric between the plates increases the capacitor’s ability to store charge for the same voltage by a factor equal to the dielectric constant, κ. For a parallel-plate capacitor with vacuum, the capacitance is C0 = ε0 A / d. When a dielectric with relative permittivity κ fills the space, the permittivity becomes ε = ε0 κ, so the capacitance becomes C = ε A / d = ε0 κ A / d = κ C0. This means the new capacitance is larger by κ, not smaller or the same, and not κ squared. The dielectric reduces the electric field for the same charge, and that reduction directly translates into a larger capacitance by the factor κ.

The key idea is that inserting a dielectric between the plates increases the capacitor’s ability to store charge for the same voltage by a factor equal to the dielectric constant, κ. For a parallel-plate capacitor with vacuum, the capacitance is C0 = ε0 A / d. When a dielectric with relative permittivity κ fills the space, the permittivity becomes ε = ε0 κ, so the capacitance becomes C = ε A / d = ε0 κ A / d = κ C0. This means the new capacitance is larger by κ, not smaller or the same, and not κ squared. The dielectric reduces the electric field for the same charge, and that reduction directly translates into a larger capacitance by the factor κ.